James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

SECTION 7.8 Improper Integrals 531
We summarize the result of Example 4 for future reference:
2 y`
1
1
p
dx is convergent if p . 1 and divergent if p < 1.
x
y
0 a
FIGURE 7
y=ƒ
x=b
t b x
Type 2: Discontinuous Integrands
Suppose that f is a positive continuous function defined on a finite interval fa, bd but
has a vertical asymptote at b. Let S be the unbounded region under the graph of f and
above the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely
in a horizontal direction. Here the region is infinite in a vertical direction.) The area of the
part of S between a and t (the shaded region in Figure 7) is
Astd − y t
f sxd dx
If it happens that Astd approaches a definite number A as t l b 2 , then we say that
the area of the region S is A and we write
a
y b
f sxd dx − lim y t
f sxd dx
a
t l b 2 a
We use this equation to define an improper integral of Type 2 even when f is not a positive
function, no matter what type of discontinuity f has at b.
Parts (b) and (c) of Definition 3 are
illustrated in Figures 8 and 9 for the
case where f sxd > 0 and f has vertical
asymptotes at a and c, respectively.
y
3 Definition of an Improper Integral of Type 2
(a) If f is continuous on fa, bd and is discontinuous at b, then
y b
f sxd dx − lim y t
f sxd dx
a
t l b 2 a
if this limit exists (as a finite number).
(b) If f is continuous on sa, bg and is discontinuous at a, then
y b
f sxd dx − lim
a
t l a 1
y b
f sxd dx
t
0 at
FIGURE 8
y
b
x
if this limit exists (as a finite number).
The improper integral y b a
f sxd dx is called convergent if the corresponding limit
exists and divergent if the limit does not exist.
(c) If f has a discontinuity at c, where a , c , b, and both y c f sxd dx and
a
f sxd dx are convergent, then we define
y b c
y b
f sxd dx − y c
f sxd dx 1 y b
f sxd dx
a
a
c
0
FIGURE 9
a c b
x
Example 5 Find y 5 1
dx.
2 sx 2 2
SOLUTION We note first that the given integral is improper because f sxd − 1ysx 2 2
has the vertical asymptote x − 2. Since the infinite discontinuity occurs at the left
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